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In a right-angled triangle with sides $a$ and $b$ and hypotenuse $c$, the altitude drawn on the hypotenuse is $x$. Prove that $ab\ =\ cx$.
"
Given:
In a right-angled triangle with sides $a$ and $b$ and hypotenuse $c$, the altitude drawn on the hypotenuse is $x$.
To do:
We have to prove that $ab\ =\ cx$.
Solution:
In the given figure, $\vartriangle ABC$ is a right angle triangle having sides $a$ and $b$ and hypotenuse $c$.
Let $BD$ be the altitude drawn on the hypotenuse $AC$.
In $\vartriangle ACB$ and $\vartriangle CDB$,
$\angle B = \angle B$ (Common)
$\angle ACB = \angle CDB = 90^o$
Therefore,
$\vartriangle ACB ∼ \vartriangle CDB$ (By AA similarity)
Hence,
$\frac{AB}{BD} = \frac{AC}{BC}$ (Corresponding parts of similar triangles are proportional)
$\frac{a}{x} = \frac{c}{b}$
$ab = cx$
Hence proved.
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